3.1 The evolution equation
In this section, let
\begin{array}{c}{C}_{T}=\{xR\mapsto {R}^{N}\text{is continuous},x(t+T)=x(t)\},\hfill \\ {W}^{1,2}=\{x\in {C}_{T}:{\int}_{0}^{T}({x}^{2}+{\dot{x}(t)}^{2})\phantom{\rule{0.2em}{0ex}}dt\mathrm{\infty}\},\hfill \end{array}
where \dot{x} is the weak derivative of x. {C}_{T} is a Banach space under the norm {\parallel x\parallel}_{c}={max}_{t\in R}x. Equipped with the norm
{\parallel x\parallel}_{1,2}={\left({\int}_{0}^{T}({x}^{2}+{\dot{x}(t)}^{2})\phantom{\rule{0.2em}{0ex}}dt\right)}^{\frac{1}{2}},
{W}^{1,2} becomes a separable Banach space. The following is our main result of this part.
Theorem 3.1 Assume the following hold:

(i)
f(t+T)=f(t) and A(t+T,x)=A(t,x) for all (t,x)\in R\times {R}^{N};

(ii)
t\to A(t,x) is measurable and f\in {L}^{2}([0,T];{R}^{N});

(iii)
for each t\in R, the operator A(t,\cdot ):{R}^{N}\to {R}^{N} is uniformly monotone and hemicontinuous, that is, there exists a constant p>0 such that (A(t,{x}_{1})A(t,{x}_{2}),{x}_{1}{x}_{2})\ge p{{x}_{1}{x}_{2}}^{2} for all {x}_{1},{x}_{2}\in {R}^{N}, and the map s\to (A(t,x+sz),y) is continuous on [0,1] for all x,y,z\in {R}^{N};

(iv)
B:{R}^{N}\to {R}^{N}
is a bounded linear operator and there exists
c\in {R}^{+}
such that
(Bx,x)\ge c{x}^{2},\phantom{\rule{1em}{0ex}}\mathrm{\forall}x\in {R}^{N},
then the problem (1.2) has a unique Tantiperiodic solution.
In order to complete the proof of Theorem 3.1, we need the following lemmas.
Lemma 3.1 (see [33])
Suppose Γ is a bounded open set of a normal space X, f is compact in \overline{\mathrm{\Gamma}} and p\in X\mathrm{\setminus}f(\partial \mathrm{\Gamma}). Then the equation f(x)=p has at least one solution in Γ, provided with deg(f,\mathrm{\Gamma},p)\ne 0.
Lemma 3.2
Consider the equation
\dot{x}+Bx=f(t)\phantom{\rule{1em}{0ex}}\mathit{\text{a.e.}}t\in R,
(3.1)
where B:{R}^{N}\to {R}^{N} is a bounded linear operator, and there exists c\in {R}^{+} such that (Bx,x)\ge c{x}^{2} for all x\in {R}^{N}, f(t+T)=f(t) and f\in {L}^{2}([0,T];{R}^{N}). Then the problem (3.1) has a unique Tantiperiodic solution.
Proof Let x be a solution of (3.1) satisfying the boundary value condition x(0)=x(T). Then x is a Tantiperiodic solution of (3.1). Denote Lx=\dot{x}+Bx for all x\in {W}^{1,2}, then L:{W}^{1,2}\to {L}^{2}([0,T];{R}^{N}) is a linear operator.
Firstly, we show L:{W}^{1,2}\to {L}^{2}([0,T];{R}^{N}) is one to one. Suppose L({x}_{1})=L({x}_{2}), then {\dot{x}}_{1}+B{x}_{1}={\dot{x}}_{2}+B{x}_{2} a.e. t\in R, and so {\dot{x}}_{1}{\dot{x}}_{2}+B{x}_{1}B{x}_{2}=0 a.e. t\in R. Take an inner product above with {x}_{1}{x}_{2} and note that
({\dot{x}}_{1}{\dot{x}}_{2},{x}_{1}{x}_{2})+(B{x}_{1}B{x}_{2},{x}_{1}{x}_{2})=0.
By using integration from 0 to T and the relation x(0)=x(T), one can see that
{\int}_{0}^{T}({\dot{x}}_{1}{\dot{x}}_{2},{x}_{1}{x}_{2})\phantom{\rule{0.2em}{0ex}}dt=0.
Since B is a linear bounded operator, then
(B{x}_{1}B{x}_{2},{x}_{1}{x}_{2})\ge c{{x}_{1}{x}_{2}}^{2}
for some constant c>0. It follows that
\begin{array}{rcl}0& =& {\int}_{0}^{T}(B{x}_{1}B{x}_{2},{x}_{1}{x}_{2})\phantom{\rule{0.2em}{0ex}}dt\\ \ge & {\int}_{0}^{T}c{{x}_{1}{x}_{2}}^{2}\phantom{\rule{0.2em}{0ex}}dt\\ =& c{\parallel {x}_{1}{x}_{2}\parallel}_{2}^{2}\\ \ge & 0.\end{array}
Hence, {x}_{1}={x}_{2} a.e. t\in R.
Next, we claim that L:{W}^{1,2}\to {L}^{2}([0,T];{R}^{N}) is surjective. For this purpose, consider the Cauchy problem
\{\begin{array}{c}\dot{x}+Bx=f(t),\hfill \\ x(0)=\eta .\hfill \end{array}
(3.2)
It is well known that the above problem has a unique solution which can be written as follows:
x(t)={e}^{Bt}\eta +{\int}_{0}^{t}{e}^{B(ts)}f(s)\phantom{\rule{0.2em}{0ex}}ds.
Since x(0)=x(T), then we have that
\eta ={e}^{BT}\eta +{\int}_{0}^{T}{e}^{B(Ts)}f(s)\phantom{\rule{0.2em}{0ex}}ds.
By hypothesis (iv), one has that {(I{e}^{BT})}^{1} exists; therefore, when we take
\eta ={(I{e}^{BT})}^{1}\times {\int}_{0}^{t}{e}^{B(ts)}f(s)\phantom{\rule{0.2em}{0ex}}ds,
the solution of the problem (3.2) is an antiperiodic solution of the problem (3.1). This completes the proof. □
Proof of Theorem 3.1 Consider the homotopic systems of (1.2),
\dot{x}+Bx=\lambda f(t)\lambda A(t,x),
(3.3)
where \lambda \in [0,1]. Obviously, \lambda f(t)\lambda A(t,x) is hemicontinuous.
First, a priori bound of the solution set is derived. We claim that there is a priori bound in {W}^{1,2} for the possible solution x(t) of (3.3). Take the inner product with x(t), and then integrate from 0 to T. It follows that
{\int}_{0}^{T}(\dot{x},x)\phantom{\rule{0.2em}{0ex}}dt+{\int}_{0}^{T}(Bx,x)\phantom{\rule{0.2em}{0ex}}dt=\lambda {\int}_{0}^{T}(f,x)\phantom{\rule{0.2em}{0ex}}dt\lambda {\int}_{0}^{T}(A(t,x),x)\phantom{\rule{0.2em}{0ex}}dt.
Without loss of generality, we assume A(t,0)=0. Since {\int}_{0}^{T}(\dot{x},x)\phantom{\rule{0.2em}{0ex}}dt=0, and then
{\int}_{0}^{T}(Bx,x)\phantom{\rule{0.2em}{0ex}}dt=\lambda {\int}_{0}^{T}(f,x)\phantom{\rule{0.2em}{0ex}}dt\lambda {\int}_{0}^{T}(A(t,x),x)\phantom{\rule{0.2em}{0ex}}dt.
By hypothesis (iii), we deduce that
c{\parallel x\parallel}_{2}^{2}\le \lambda {\parallel f\parallel}_{2}{\parallel x\parallel}_{2},
which implies
{\parallel x\parallel}_{2}\le {M}_{1}
(3.4)
for some constant {M}_{1}>0. Hence, there is a constant \tau \in [0,T] such that
{x(\tau )}^{2}\le {M}_{2}
(3.5)
for some constant {M}_{2}>0. By (3.3), one has that
\begin{array}{rcl}(\dot{x},x)& =& \lambda (f,x)\lambda (A(t,x),x)(Bx,x)\\ \le & \lambda f\cdot x\lambda p{x}^{2}.\end{array}
Integrating above from τ to t, we have that
{\int}_{\tau}^{t}(\dot{x},x)\phantom{\rule{0.2em}{0ex}}dt\le {\int}_{0}^{T}f\cdot x\phantom{\rule{0.2em}{0ex}}dt\le {\parallel f\parallel}_{2}{\parallel x\parallel}_{2}.
From {\int}_{\tau}^{t}(\dot{x},x)\phantom{\rule{0.2em}{0ex}}dt={x(t)}^{2}{x(\tau )}^{2}, we know
{x(t)}^{2}{x(\tau )}^{2}\le {\parallel f\parallel}_{2}{\parallel x\parallel}_{2}.
(3.6)
By (3.4) and (3.5), we obtain that there is some constant {M}_{3}>0 (independent of λ) such that
x(t)\le {({x(\tau )}^{2}+{\parallel f\parallel}_{2}{\parallel x\parallel}_{2})}^{\frac{1}{2}}\le {M}_{3}
for any t\in R. Thus,
{\parallel x\parallel}_{c}=\underset{t\in R}{max}x(t)\le {M}_{3}.
(3.7)
Since the operator A is hemicontinuous, and B is a bounded linear operator, we show that
\begin{array}{c}{\parallel \dot{x}\parallel}_{2}\le {M}_{4},\hfill \\ {\parallel x\parallel}_{1,2}\le {M}_{5},\hfill \end{array}
where the constants {M}_{4},{M}_{5}>0. The claim is proved.
Secondly, we can prove the existence of antiperiodic solutions for Eq. (1.2). Set
\mathrm{\Gamma}=\{x\in {W}^{1,2}:{\parallel x\parallel}_{1,2}<{M}_{5}+1\}.
Then Γ is a bounded open set in {W}^{1,2}. By Lemma 3.2, it is easy to see that
{L}^{1}:{L}^{2}([0,T];{R}^{N})\to {W}^{1,2}
is well defined. We define the operator N:\mathrm{\Gamma}\to {W}^{1,2}, N(x)={L}^{1}(fA(t,x)). Obviously, N is compact. Hence, the fixed point of N in \overline{\mathrm{\Gamma}} is the antiperiodic solutions of Eq. (1.2). Let {h}_{\lambda}(x):\overline{\mathrm{\Gamma}}\times [0,1]\to {W}^{1,2}
{h}_{\lambda}(x)=x\lambda N(x).
By (3.7), we obtain \theta \phantom{\rule{0.2em}{0ex}}\overline{\in}\phantom{\rule{0.2em}{0ex}}h(\partial \mathrm{\Gamma}). So for each \lambda \in [0,1], then we have that
\begin{array}{rcl}deg({h}_{\lambda},\mathrm{\Gamma},\theta )& =& deg({h}_{1},\mathrm{\Gamma},\theta )\\ =& deg(\mathit{id}N,\mathrm{\Gamma},\theta )\\ =& deg({h}_{0},\mathrm{\Gamma},\theta )\\ =& deg(\mathit{id},\mathrm{\Gamma},\theta )=1,\end{array}
where id is the identity. Consequently, N has a fixed point in Γ by Lemma 3.1. Namely, Eq. (1.2) has an antiperiodic solution.
Next, we prove the uniqueness. Suppose that {x}_{1}, {x}_{2} are two solutions of Eq. (1.2). Then
{\dot{x}}_{1}+A(t,{x}_{1})+B{x}_{1}={\dot{x}}_{2}+A(t,{x}_{2})+B{x}_{2}.
So,
{\dot{x}}_{1}{\dot{x}}_{2}+A(t,{x}_{1})A(t,{x}_{2})+B{x}_{1}B{x}_{2}=0.
Take an inner product above with {x}_{1}{x}_{2} and note that
({\dot{x}}_{1}{\dot{x}}_{2},{x}_{1}{x}_{2})+(A(t,{x}_{1})A(t,{x}_{2}),{x}_{1}{x}_{2})+(B{x}_{1}B{x}_{2},{x}_{1}{x}_{2})=0.
By using integration from 0 to T and the relation x(0)=x(T), one can see that
\begin{array}{rcl}0& =& {\int}_{0}^{T}(A(t,{x}_{1})A(t,{x}_{2}),{x}_{1}{x}_{2})\phantom{\rule{0.2em}{0ex}}dt+{\int}_{0}^{T}(B{x}_{1}B{x}_{2},{x}_{1}{x}_{2})\phantom{\rule{0.2em}{0ex}}dt\\ \ge & {\int}_{0}^{T}(c+p){{x}_{1}{x}_{2}}^{2}\phantom{\rule{0.2em}{0ex}}dt\\ =& (c+p){\parallel {x}_{1}{x}_{2}\parallel}_{2}^{2}\\ \ge & 0.\end{array}
Hence, {x}_{1}={x}_{2} a.e. t\in R. This ends the proof. □
3.2 The evolution inclusions
Let I=[0,T] and C(I;{R}^{N}) be all the continuous functions from I to {R}^{N} with the max norm. Let {C}_{\beta}=\{v(\cdot )\in C(I;{R}^{N}):v(0)=v(T)\}, and {W}^{1,2}(I,{R}^{N})=\{u(\cdot )\in {C}_{\beta}:\dot{u}(\cdot )\in {L}^{2}(I;{R}^{N})\}. {W}^{1,2}(I,{R}^{N}) is a separable Banach space under the norm {\parallel \cdot \parallel}_{1,2}.
Consider the following antiperiodic problem:
\{\begin{array}{c}\dot{x}+A(t,x)+Bx\in F(t,x)\phantom{\rule{1em}{0ex}}\text{a.e.}t\in I,\hfill \\ x(T)=x(0),\hfill \end{array}
(3.8)
where A:{R}^{N}\to {R}^{N} is a hemicontinuous function, B is a bounded linear operator from {R}^{N} to {R}^{N}, and F:R\times {R}^{N}\to {2}^{{R}^{N}} is a multifunction. By a solution x of the problem (3.8), we mean a function x\in {W}^{1,2}(I,{R}^{N}), and there exists a function f(t)\in F(t,x(t)) such that
\u3008\dot{x}(t),v\u3009+\u3008A(t,x(t)),v\u3009+\u3008Bx,v\u3009=\u3008f(t),v\u3009
for all v\in {R}^{N} and almost all t\in I.
In this section, we prove two existence theorems under the hypothesis that the multivalued nonlinearity F is convexvalued (‘convex existence theorem’) or nonconvexvalued (‘nonconvex existence theorem’). The precise hypotheses on the data of the problem (3.8) are as follows:
H(A): A:I\times {R}^{N}\to {R}^{N} is a nonlinear function such that

(i)
t\to A(t,x) is measurable;

(ii)
for each t\in I, the operator A(t,\cdot ):{R}^{N}\to {R}^{N} is uniformly monotone and hemicontinuous, that is, there exists a constant p>0 such that (A(t,{x}_{1})A(t,{x}_{2}),{x}_{1}{x}_{2})\ge p{{x}_{1}{x}_{2}}^{2} for all {x}_{1},{x}_{2}\in {R}^{N}, and the map s\mapsto (A(t,x+sz),y) is continuous on [0,1] for all x,y,z\in {R}^{N}.
H(B): B:{R}^{N}\to {R}^{N} is a bounded linear operator, and there exists c\in {R}^{+} such that
(Bx,x)\ge c{x}^{2},\phantom{\rule{1em}{0ex}}\mathrm{\forall}x\in {R}^{N}.
H{(F)}_{1}: F:R\times {R}^{N}\to {P}_{k}({R}^{N}) is a multifunction such that

(i)
(t,x)\to F(t,x) is graph measurable;

(ii)
for almost all t\in I, x\to F(t,x) is LSC;

(iii)
there exists an nonnegative function b(\cdot )\in {L}_{+}^{2}(I) and a constant {c}_{1}>0 such that
F(t,x)=sup\{\parallel f\parallel :f\in F(t,x)\}\le b(t)+{c}_{1}{x}^{\alpha},
for all x\in {R}^{N}, t\in T, where \alpha <1 or \alpha =1 with {c}_{1}<c (c in H(B)).
H{(F)}_{2}: F:I\times {R}^{N}\to {P}_{kc}({R}^{N}) is a multifunction such that

(i)
(t,x)\to F(t,x) is graph measurable;

(ii)
for almost all t\in I, x\to F(t,x) has a closed graph; and H{(F)}_{1}(iii) holds.
Theorem 3.2 If hypotheses H(A), H(B) and H{(F)}_{1} hold, then the problem (3.8) has a solution x\in {W}^{1,2}(I,{R}^{N}).
Proof Let Lx=\dot{x}+A(t,x)+Bx for all x\in {W}^{1,2}(I,{R}^{N}). By Theorem 3.1, we have L:{W}^{1,2}(I,{R}^{N})\to {L}^{2}([0,T];{R}^{N}) is one to one and surjective, and so {L}^{1}:{L}^{2}([0,T];{R}^{N})\to {W}^{1,2}(I,{R}^{N}) is well defined. So, we prove that
{L}^{1}:{L}^{2}([0,T];{R}^{N})\to {L}^{2}([0,T];{R}^{N})({W}^{1,2}(I,{R}^{N})\subset {L}^{2}([0,T];{R}^{N}))
is completely continuous (i.e., it is continuous and maps bounded sets into relatively compact sets). To this end, let K\subset {L}^{2}([0,T];{R}^{N}) be bounded. We shall show that {L}^{1}(K) is relatively compact in {L}^{2}([0,T];{R}^{N}). For this purpose, let x\in {L}^{1}(K), then x={L}^{1}(u) with u\in K. By (3.7), we have {\parallel x\parallel}_{2}\le c{\parallel Lx\parallel}_{2}={\parallel u\parallel}_{2}\le cK=csup\{{\parallel u\parallel}_{2}:u\in K\}<+\mathrm{\infty} and {\parallel \dot{x}\parallel}_{2}\le {\parallel u\parallel}_{2}+{\parallel A(x)\parallel}_{2}+{\parallel Bx\parallel}_{2}\le M for some constant M>0. From these bounds we infer that {L}^{1}(K) is bounded in {W}^{1,2}(I,{R}^{N}). But {W}^{1,2}(I,{R}^{N}) is compactly embedded in {L}^{2}([0,T];{R}^{N}). Therefore, {L}^{1}(K) is relatively compact in {L}^{2}([0,T];{R}^{N}). Also, from the fact that {L}^{1} is a compact operator, {L}^{1}:{L}^{2}([0,T];{R}^{N})\to {L}^{2}([0,T];{R}^{N}) is continuous.
Next, let N:{L}^{2}([0,T];{R}^{N})\to {2}^{{L}^{2}([0,T];{R}^{N})} be the multivalued Nemitsky operator corresponding to F and N be defined by N(x)=\{v\in {L}^{2}([0,T];{R}^{N}):v(t)\in F(t,x(t))\} a.e. on I.
We claim that N(\cdot ) has nonempty, closed, decomposable values and is LSC. The closedness and decomposability of the values of N(\cdot ) are easy to check. For the nonemptiness, note that if x\in {L}^{2}([0,T];{R}^{N}), by hypothesis H{(F)}_{1}(i), (t,x)\to F(t,x) is graph measurable, so we apply Aumann’s selection theorem and obtain a measurable map v:I\to {R}^{N} such that v(t)\in F(t,x(t)) a.e. on I. By hypothesis H{(F)}_{1}(iii), v\in {L}^{2}([0,T];{R}^{N}). Thus for every x\in {R}^{N}, N(x)\ne \mathrm{\varnothing}. To prove the lower semicontinuity of N(\cdot ), we only show that every u\in {L}^{2}([0,T];{R}^{N}), x\to d(u,N(x)) is a USC {R}_{+}valued function. Note that
\begin{array}{rcl}d(u,N(x))& =& inf\{{\parallel uv\parallel}_{2}:v\in N(x)\}\\ =& inf\{{\left({\int}_{0}^{T}{u(t)v(t)}^{2}\phantom{\rule{0.2em}{0ex}}dt\right)}^{\frac{1}{2}}:v\in N(x)\}\\ =& {({\int}_{0}^{T}inf\{{u(t)v(t)}_{2}:v\in F(t,x)\}\phantom{\rule{0.2em}{0ex}}dt)}^{\frac{1}{2}}\\ =& {\left({\int}_{0}^{T}d{(u(t),F(t,x))}^{2}\phantom{\rule{0.2em}{0ex}}dt\right)}^{\frac{1}{2}}\end{array}
(see Hiai and Umegaki [34] Th. 2.2). We shall show that for every \lambda \ge 0, the superlevel set {U}_{\lambda}=\{x\in {L}^{2}([0,T];{R}^{N}):d(u,N(x))\ge \lambda \} is closed in {L}^{2}([0,T];{R}^{N}). Let {\{{x}_{n}\}}_{n\ge 1}\subseteq {U}_{\lambda} and assume that {x}_{n}\to x in {L}^{2}([0,T];{R}^{N}). By passing to a subsequence, if necessary, we may assume that {x}_{n}(t)\to x(t) a.e. on I as n\to \mathrm{\infty}. By hypothesis H{(F)}_{1}(ii), x\to d(u,F(t,x)) is an upper semicontinuous {R}_{+}valued function. So, via Fatou’s lemma, we have
\begin{array}{rcl}{\lambda}^{2}\le \overline{lim}{\left[d(u,N({x}_{n}))\right]}^{2}& =& \overline{lim}{\int}_{0}^{T}{\left[d(u(t),F(t,{x}_{n}))\right]}^{2}\phantom{\rule{0.2em}{0ex}}dx\\ \le & {\int}_{0}^{T}\overline{lim}{\left[d(u(t),F(t,{x}_{n}))\right]}^{2}\phantom{\rule{0.2em}{0ex}}dx\\ \le & {\int}_{0}^{T}{\left[d(u,F(t,x))\right]}^{2}\phantom{\rule{0.2em}{0ex}}dt={\left[d(u,N(x))\right]}^{2}.\end{array}
Therefore, x\in {U}_{\lambda} and this proves the LSC of N(\cdot ).
We apply Lemma 2.2 and obtain a continuous map f:{L}^{2}([0,T];{R}^{N})\to {L}^{2}([0,T];{R}^{N}) such that f(x)\in N(x). To finish our proof, we only need to solve the fixed point problem: x={L}^{1}f(x).
We claim that the set \mathrm{\Gamma}=\{x\in {L}^{2}([0,T];{R}^{N}):x=\sigma {L}^{1}f(x),\sigma \in (0,1)\} is bounded. Let x\in \mathrm{\Gamma}, then x=\sigma {L}^{1}f(x). By hypothesis H{(F)}_{1}(iii), we can derive
f(x)\le b(t)+{c}_{1}{x}^{\alpha},
then
\begin{array}{rcl}{\parallel f(x)\parallel}_{2}& \le & {\parallel b\parallel}_{2}+{\parallel {c}_{1}{x}^{\alpha}\parallel}_{2}\\ \le & {\parallel b\parallel}_{2}+{\left({\parallel {c}_{1}^{2}\parallel}_{\frac{1}{1\alpha}}{\parallel {x}^{2\alpha}\parallel}_{\frac{1}{\alpha}}\right)}^{1/2}\\ =& {\parallel b\parallel}_{2}+{\parallel {c}_{1}\parallel}_{\frac{2}{1\alpha}}{\parallel x\parallel}_{2}^{\alpha}\\ =& {\parallel b\parallel}_{2}+{c}_{1}{T}^{\frac{2}{1\alpha}}{\parallel x\parallel}_{2}^{\alpha}\end{array}
with \alpha <1. By (3.6), we get that
{\parallel x\parallel}_{2}\le {c}_{2}{\parallel f\parallel}_{2},
for some constant {c}_{2}>0. So, we have that
{\parallel x\parallel}_{2}\le {c}_{2}{\parallel b\parallel}_{2}+{c}_{1}{c}_{2}{T}^{\frac{2}{1\alpha}}{\parallel x\parallel}_{2}^{\alpha}.
Thus, we can find a constant {c}_{3}>0 such that {\parallel x\parallel}_{2}\le {c}_{3}. If \alpha =1, we can also find a constant \overline{{c}_{3}}=\frac{{\parallel b\parallel}_{2}}{c{c}_{1}}>0 such that {\parallel x\parallel}_{2}\le \overline{{c}_{3}}. Similar to the estimation of (3.7), we have that
{\parallel x\parallel}_{c}=\underset{t\in R}{max}x(t)\le {c}_{4},
(3.9)
for some constant {c}_{4}>0. So, Γ is bounded in {L}^{2}([0,T];{R}^{N}). Invoking LeraySchauder’s alternative theorem, we obtain there exists x\in {W}^{1,2}(I,{R}^{N}) such that x={L}^{1}f(x), x is a solution of the problem (3.8). This ends the proof. □
Theorem 3.3 If hypotheses H(A), H(B) and H{(F)}_{2} hold, then the problem (3.8) has a solution x\in {W}^{1,2}(I,{R}^{N}). Moreover, the solution set is weakly compact in {W}^{1,2}(I,{R}^{N}).
Proof The proof is as that of Theorem 3.2. So, we only present those particular points where the two proofs differ.
In this case, the multivalued Nemistsky operator N:{L}^{2}([0,T];{R}^{N})\to {2}^{{L}^{2}([0,T];{R}^{N})} has nonempty closed, convex values in {L}^{2}([0,T];{R}^{N}) and is USC. The closedness and convexity of the values of N(\cdot ) are clear. To prove the nonemptiness, let x\in {W}^{1,2}(I,{R}^{N}) and {\{{s}_{n}\}}_{n\ge 1} be a sequence of step functions such that
{s}_{n}(t)\to x\phantom{\rule{1em}{0ex}}\text{and}\phantom{\rule{1em}{0ex}}{s}_{n}(t)\le x(t)\phantom{\rule{1em}{0ex}}\text{a.e. on}I.
Then by virtue of hypothesis H{(F)}_{2}(i), for every n\ge 1, t\to F(t,{s}_{n}) admits a measurable selector {f}_{n}(t). From hypothesis H{(F)}_{2}(iii), we have that there exists a constant {c}_{5}>0 such that
\underset{{f}_{n}\in F(t,{s}_{n})}{sup}{\parallel {f}_{n}\parallel}_{2}\le {\parallel b\parallel}_{2}+{c}_{5}{\parallel x\parallel}_{2}^{\alpha}.
So {\{{f}_{n}\}}_{n\ge 1} is uniformly integrable. By the DunfordPettis theorem, and by passing to a subsequence if necessary, we may assume that {f}_{n}\to f weakly in {L}^{2}([0,T];{R}^{N}). Then from Theorem 3.1 in [35], we have
the last inclusion being a consequence of hypothesis H{(F)}_{2}(ii). So f\in N(x). Thus we prove the nonemptiness of N(\cdot ).
Next, we show that N(\cdot ) is USC from {W}^{1,2}(I,{R}^{N}) into {L}^{2}{([0,T];{R}^{N})}_{w}. Let C be a nonempty and weakly closed subset of {L}^{2}([0,T];{R}^{N}). We need to show that the set
{N}^{1}(C)=\{x\in D(L):N(x)\cap C\ne \varphi \}
is closed. Let {\{{x}_{n}\}}_{n\ge 1}\subseteq {N}^{1}(C) and assume {x}_{n}\to x in {W}^{1,2}(I,{R}^{N}). Passing to a subsequence, we can get that {x}_{n}(t)\to x(t) a.e. on I. Let {f}_{n}\in N({x}_{n})\cap C, n\ge 1. Then by virtue of hypothesis H{(F)}_{2}(iii) and the DunfordPettis theorem, we may assume that {f}_{n}\to f\in C weakly in {L}^{2}([0,T];{R}^{N}). As before, we have
then f\in N(x)\cap C, i.e., {N}^{}(C) is closed in {W}^{1,2}(I,{R}^{N}). This proves the upper semicontinuity of N(\cdot ) from {W}^{1,2}(I,{R}^{N}) into {L}^{2}{([0,T];{R}^{N})}_{w}.
We consider the following fixed point problem:
Recalling that {L}^{1}:{L}^{2}([0,T];{R}^{N})\to {L}^{2}([0,T];{R}^{N}) is completely continuous, we see that {L}^{1}N:{L}^{2}([0,T];{R}^{N})\to {P}_{kc}({L}^{2}([0,T];{R}^{N})) is USC and maps bounded sets into relatively compact sets. We easily check that the set
{\mathrm{\Gamma}}_{1}=\{u\in {L}^{2}([0,T];{R}^{N}):u\in \sigma {L}^{1}N(u),\sigma \in (0,1)\}
is bounded, as in the proof of Theorem 3.2. Invoking Lemma 2.1, there exists u\in {W}^{1,2}(I,{R}^{N}) such that u\in {L}^{1}N(u). Evidently, this is a solution of the problem (3.8).
Let S denote the solution set of the problem (3.8). As in the proof of Theorem 3.2, we have that S=sup\{{\parallel u\parallel}_{1,2}:u\in S\}\le M, where M>0. By virtue of hypothesis H{(F)}_{2}(iii) and the DunfordPettis theorem, we may assume that {u}_{n}\to u weakly in {W}^{1,2}(I,{R}^{N}). As before, we have
then u\in S, hence S is weakly compact in {W}^{1,2}(I,{R}^{N}). □